jquery - 使用 jQuery 和 Ajax 提交 Rails 表单

Question

编辑:想通了，所以问了一个相关的问题。

这是我的 JavaScript

jQuery.ajaxSetup({
    'beforeSend': function(xhr) {
        xhr.setRequestHeader("Accept", "text/javascript")
    }
})

jQuery.fn.submitWithAjax = function() {
    this.submit(function() {
        $.post(this.action, $(this).serialize(), null, "script");
        return false;
    })
    return this;
};

    $('.error').hide();
$("#business_submit").click(function() {
    // validate and process form here

    $('.error').hide();
    var name = $("input#business_name").val();
    if (name == ""  || name == "Required Field") {
        $('#namelabel').show()
        $("#business_name").focus();
        return false;
    }
    var address = $("#business_address").val();
    if (address == ""  || address == "Required Field") {
        $('#addresslabel').show();
        $("#business_address").focus();
        return false;
    }
    var city = $("#business_city").val();
    if (city == "" || city == "Required Field") {
        $('#citylabel').show();
        $('#business_city').focus();
        return false;
    }
    var state = $("#business_state").val();
    if (state == ""  || state == "Required Field") {
        $('#statelabel').show();
        $("#business_state").focus();
        return false;
    }
    var zip = $("#business_zip").val();
    if (zip == ""  || zip == "Required Field") {
        $('#ziplabel').show();
        $("#business_zip").focus();
        return false;
    }
    var phone = $("#business_phone").val();
    if (phone == ""  || phone == "Required Field") {
        $('#phonelabel').show();
        $("#business_phone").focus();
        return false;
    }

    var category = $("#business_business_category_id").val();
    if (category == " - Choose one - ") {
        $('#categorylabel').show();
        $("#business_business_category_id").focus();
        return false;
    }


   $.ajax ({
        type: "POST",
        data: form.serialize()
    });
    return false;
})

.ajax 代码触发我的 create.js.erb 文件，其中包含

$("#new_business").submitWithAjax();
$("#new_business")[0].reset();
$("#new_business").hide();

这是表单下方的表格。

<div id="unapproved">
  <table width="650" align="center" cellpadding="6" cellspacing="0">

    <tr>
      <td width="150"><a class="Contact<%=h @business.id %>" href="#"><%=h @business.name %></a></td>
      <td width="150"><%=h @business.address %></td>
      <td width="100"><%=h @business.business_category.name %></td>
      <td width="200"><%=h @business.description %></td>
      <td width="50"><%= link_to(image_tag('/images/accept.png', :alt => 'Approve'), :id =>@business.id, :action => 'approve') %>
        <a class="Edit<%=h @business.id %>" href="#"><img src="/images/pencil.png" alt="Edit" /></a>
      <%= link_to(image_tag('/images/bin.png', :alt => 'Remove'), @business, :confirm => 'Are you sure?', :method => :delete) %></td>
    </tr>
  </table>
</div>

现在我唯一的问题是我的表单下面的表格没有动态刷新。它将数据很好地添加到数据库中，并且所有验证都运行良好。但我必须刷新页面。我尝试添加类似的内容

$("#unapproved").append("<%= escape_javascript(render(:file => 'business')) %>");

但这只会破坏它。

Answer 1

我的猜测(因为您的“提交”按钮不在上面的 HTML 中，所以您的“business_submit”按钮只是一个输入而不是提交按钮。

这段代码:

jQuery.fn.submitWithAjax = function() {
    this.submit(function() {
        $.post(this.action, $(this).serialize(), null, "script");
        return false;
    })
    return this; };

不告诉表单提交。它说当“提交”发生时，去做某事。

From the jQuery documentation, submit(fn) is used to "Bind a function to the submit event of each matched element. The submit event fires when a form is submitted."

您实际上从未提交过表单。我愿意打赌(不能确定，因为我没有所有代码)，您将 SubmitWithAjax 方法更改为:

jQuery.fn.submitWithAjax = function() {
    $.post(this.action, $(this).serialize(), null, "script");
    return false; };

它应该可以工作。它至少更接近您想要做的事情。